If the following system of linear equations
$2x + y + z = 5$
$x - y + z = 3$
$x + y + az = b$
has no solution,then :

If $p$ and $q$ are two distinct real values of $\lambda$ for which the system of equations $\begin{aligned} (\lambda-1) x+(3 \lambda+1) y+2 \lambda z &=0 \\ (\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z &=0 \\ 2 x+(3 \lambda+1) y+3(\lambda-1) z &=0 \end{aligned}$ has a non-zero solution,then $p^2+q^2-p q=$

Let $n$ be the number obtained on rolling a fair die. If the probability that the system of equations
$x-ny+z=6$
$x+(n-2)y+(n+1)z=8$
$(n-1)y+z=1$
has a unique solution is $\frac{k}{6}$,then the sum of $k$ and all possible values of $n$ is:

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

Consider the simultaneous linear equations $\beta x + \alpha y - z = -1$,$3x - \beta y + \alpha z = 0$,and $\alpha x + \beta y + z = 1$. In the usual notation used in Cramer's rule,given that $\frac{\Delta_1}{\Delta} = -1$,$\frac{\Delta_2}{\Delta} = 1$,and $\frac{\Delta_3}{\Delta} = 2$,then $(\alpha, \beta) = $

The system of equations $x + ky - z = 0$,$3x - ky - z = 0$,and $x - 3y + z = 0$ has a non-zero solution for $k =$